# Fated to earn money [closed]

Here's a puzzle I made. Joe has investments in Company A, Company B, and Company C.

Joe is fated to earn $25.00 from Company A within 2 days from now. Joe is fated to earn$45.00 from Company B within 3 days from now.

Joe is fated to earn $100.00 from Company C within 5 days from now. Joe is fated to earn no more than$26.00 from Company C and Company B on day 1 (1 day from now).

Joe is fated to earn at least $14.00 from Company A and Company C on day 2 (2 days from now). Joe has to earn twice the amount of money on the first day than the second day from Companies A, B, and C and twice the amount of money on the second day than the third day from Companies A, B, and C. This can be expressed algebraically as Joe earning x money on day 3 (3 days from now), 2x money on day 2 (2 days from now), and 4x money on day 1 (1 day from now). Joe can earn whatever amount of money (that satisfies the other conditions) from Companies A, B, and C on day 4 and day 5 (4 and 5 days from now). What is the lowest amount of money Joe can earn on day 1 (1 day from now) from Companies A, B, and C? Explain your reasoning. P.S. How come there doesn't seem to be good formulas to use for this question? Hint: You won't have to calculate in cents for this question. • Is that 4x money on day 1? – kushj Sep 10 '19 at 17:23 • Yes, it is, my bad. Edited. Copied my question wrong. Should be ok now. – Yukang Jiang Sep 10 '19 at 17:24 • When you say within earn X money within 2 days, is it valid for just first 2 days, or any pair of consecutive two days? – kushj Sep 10 '19 at 17:25 • This appears to be a straightforward mathematics question. I can't see the puzzley/fun part of it? – Arnaud Mortier Sep 10 '19 at 17:31 • @YukangJiang No, MSE is not usually for questions where you already know the answer. If there is a trick here to avoid algebra, then it's perfectly fine. – Arnaud Mortier Sep 10 '19 at 17:39 ## 1 Answer So I started with hard limit, and tried to adjust the numbers, which was pretty straight forward, There may be other permutations possible, haven't really gotten into algebra portion of it. Since at least A and B earnings together are 70 dollars over first 3 days, then it means that best outcome is: 40/20/10. (4x + 2x + x = 70) At same time, Day 1 maximum income is 51 dollars (A: 25 dollars, B+C: 26 dollars). Since, it is mentioned answer is not in cents, then only solution space is 40, 44, and 48. So brute force solution: (Start by keeping in mind that C earning can be adjusted for Day 4 and day 5, so only add it when required) -- Fix the maximum earning for B on day 1, and adjust for A. -- This gives Earning for A on day 2. That gives the earning for C on day 2, which gives the earning for B -- Day 3 is pretty simple, adjust for B, and check if equation makes sense. If it does, answer achieved, else move on to next Try for 40 Day 1: 14/26/0 Day 2: 11/6/3 Day 3: 0/10*/0 B required, 13, not possible Try for 44: Day 1: 18/26/0 Day 2: 7/8/7 Day 3: 0/11/0 And voila :) • For Company A, you have 14 on day 1 and 9 on day 2. That's 23 dollars total not 25 dollars, but you are right in C earnings only being necessary to add when required – Yukang Jiang Sep 10 '19 at 18:00 • Fixed, thankfully didn't changed the answer – kushj Sep 10 '19 at 18:04 • You have total Company B earnings at 26 (Day 1) + 6 (Day 2) + 8 (Day 3). That's 40 dollars, not 45. Company B earnings should be 45 dollars over the three day or less period. I think the answer is a little higher than 40 dollars for day 1. – Yukang Jiang Sep 10 '19 at 18:09 • Ah my bad.. somehow, while solving$40 got stuck in my head for B, and I didn't double check :\ – kushj Sep 10 '19 at 18:11
• Brute force solution. Hopefully, someone will have more elegant solution, and even more hopefully, my sleep deprived brain haven't made even more silly errors this time – kushj Sep 10 '19 at 18:21